Final answer:
The maximum mass of AgCl that can precipitate when 10.0 mL of a 0.00700 M solution of Cl- ions reacts with an excess of a 0.500 M solution of AgNO3 is 0.01003 grams.
Step-by-step explanation:
To calculate the maximum mass of AgCl that precipitates, we start by determining the limiting reactant. Since precipitation reactions occur in a stoichiometric ratio, and the reaction between Ag+ ions (from AgNO3) and Cl- ions (from a chloride solution such as NaCl) to form AgCl is 1:1, we need to compare the moles of each reactant available.
- First, calculate the moles of Cl- ions: (10.0 mL) × (0.00700 mol/L) = 0.000070 mol Cl-.
- Next, determine the moles of Ag+: since this is in excess, we use the moles of Cl- as the limiting factor.
- The molar mass of AgCl is 143.32 g/mol.
- To find the maximum mass of AgCl that can precipitate, multiply the moles of Cl- by the molar mass of AgCl: 0.000070 mol × 143.32 g/mol = 0.01003 g of AgCl.
Therefore, the maximum mass of AgCl that can precipitate from the given solution is 0.01003 grams.